assignment makes integer from pointer without a cast enabled by default

Converting Pointers to Integers: Avoiding Cast Errors & Mastering the Process

In this guide, we will cover how to convert pointers to integers and vice versa without running into any cast errors. We will also walk you through the step-by-step process of mastering pointer and integer conversions in C/C++.

Table of Contents

• Why Convert Pointers to Integers
• Understanding uintptr_t
• Step-by-Step Guide
• Converting Pointers to Integers
• Converting Integers to Pointers

Why Convert Pointers to Integers? {#why-convert-pointers-to-integers}

There are several use cases where you might need to convert pointers to integers and vice versa. Some common reasons include:

• Manipulating memory addresses for low-level programming.
• Serializing and deserializing data.
• Storing pointers in a generic data structure.
• Debugging and logging purposes.

However, when converting pointers to integers, it is crucial to avoid any errors that may arise from incorrect casting.

Understanding uintptr_t {#understanding-uintptr_t}

To safely convert pointers to integers, it is essential to use the uintptr_t data type. This is an unsigned integer type that is large enough to store the value of a pointer. It is available in the <stdint.h> header in C and the <cstdint> header in C++.

Using uintptr_t , you can safely cast a pointer to an integer and back to a pointer without losing any information. This ensures that the process is safe, fast, and efficient.

Step-by-Step Guide {#step-by-step-guide}

Converting pointers to integers {#converting-pointers-to-integers}.

To convert a pointer to an integer, follow these steps:

• Include the <stdint.h> header (C) or the <cstdint> header (C++) in your program.
• Cast your pointer to uintptr_t .

Converting Integers to Pointers {#converting-integers-to-pointers}

To convert an integer to a pointer, follow these steps:

• Cast your integer to the required pointer type using a double cast.

FAQs {#faqs}

Why can't i just use a regular int or unsigned int to store pointers {#regular-int}.

While it may work on some platforms where the size of an int is equal to the size of a pointer, it is not guaranteed to be portable across different systems. Using uintptr_t ensures your code remains portable and safe.

Are there performance implications when using uintptr_t ? {#performance}

The performance impact of using uintptr_t is minimal. Most modern compilers can optimize the casting operations, resulting in little to no overhead.

When should I use intptr_t instead of uintptr_t ? {#intptr_t}

intptr_t is a signed integer type that can hold a pointer value. It is useful when you need to perform arithmetic operations on pointers that may result in negative values. However, in most cases, uintptr_t is recommended.

Is it safe to perform arithmetic operations on integers representing pointers? {#pointer-arithmetic}

Performing arithmetic operations on integers representing pointers can lead to undefined behavior if the resulting integer doesn't correspond to a valid memory address. It is generally safer to perform arithmetic operations on pointers directly.

How do I avoid losing information when casting pointers to integers? {#avoid-losing-information}

By using uintptr_t , you ensure that the integer is large enough to store the value of a pointer without losing any information. Make sure always to use uintptr_t when converting pointers to integers.

Related Links

• C++ Reference: uintptr_t
• C Reference: uintptr_t
• Understanding Pointers in C and C++

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Posted on Oct 4, 2023 • Originally published at coderlegion.com

Assignment makes integer from pointer without a cast in c

Programming can be both rewarding and challenging. You work hard on your code, and just when it seems to be functioning perfectly, an error message pops up on your screen, leaving you frustrated and clueless about what went wrong. One common error that programmers encounter is the "Assignment makes integer from pointer without a cast" error in C.

This error occurs when you try to assign a value from a pointer variable to an integer variable without properly casting it. To fix this error, you need to make sure that you cast the pointer value to the appropriate data type before assigning it to an integer variable. In this article, we will dive deeper into the causes of this error and provide you with solutions to overcome it.

What makes this error occur?

I will present some cases that triggers that error to occur, and they are all have the same concept, so if you understanded why the failure happens, then you will figure out how to solve all the cases easily.

Case 1: Assignment of a pointer to an integer variable

In this simple code we have three variables, an integer pointer "ptr", and two integers "n1" and "n2". We assign 2 to "n1", so far so good, then we assign the address of "n2" to "ptr" which is the suitable storing data type for a pointer, so no problems untill now, till we get to this line "n2 = ptr" when we try to assign "ptr" which is a memory address to "n2" that needs to store an integer data type because it's not a pointer.

Case 2: Returning a Pointer from a Function that Should Return an Integer

As you can see, it's another situation but it's the same idea which causes the compilation error. We are trying here to assign the return of getinteger function which is a pointer that conatains a memory address to the result variable that is an int type which needs an integer

Case 3: Misusing Array Names as Pointers

As we might already know, that the identifier (name) of the array is actually a pointer to the array first element memory address, so it's a pointer after all, and assigning a pointer type to int type causes the same compilation error.

The solutions

The key to avoiding the error is understanding that pointers and integers are different types of variables in C. Pointers hold memory addresses, while integers hold numeric values. We can use either casting, dereferencing the pointer or just redesign another solution for the problem we are working on that allows the two types to be the same. It all depending on the situation.

Let's try to solve the above cases:

Case 1: Solution: Deferencing the pointer

We need in this case to asssign an int type to "n2" not a pointer or memory address, so how do we get the value of the variable that the pointer "ptr" pointing to? We get it by deferencing the pointer, so the code after the fix will be like the following:

Case 2: Solution: Choosing the right data type

In this case we have two options, either we change the getinteger returning type to int or change the result variable type to a pointer. I will go with the latter option, because there are a lot of functions in the C standard library that returning a pointer, so what we can control is our variable that takes the function return. So the code after the fix will be like the following:

We here changed the result variable from normal int to an int pointer by adding "*".

Case 3: Solution: Using the array subscript operator

In this case we can get the value of any number in the array by using the subscript opeartor ([]) on the array with the index number like: myarray[1] for the second element which is 2. If we still remember that the array identifier is a pointer to the array first memory, then we can also get the value of the array first element by deferencing the array identifier like: *myarray which will get us 1.

But let's solve the case by using the subscript opeartor which is the more obvious way. So the code will be like the following:

Now the number 1 is assigned to myint without any compilation erros.

The conclusion

In conclusion, the error "assignment to ‘int’ from ‘int *’ makes integer from pointer without a cast" arises in C programming when there is an attempt to assign a memory address (held by a pointer) directly to an integer variable. This is a type mismatch as pointers and integers are fundamentally different types of variables in C.

To avoid or correct this error, programmers need to ensure they are handling pointers and integers appropriately. If the intent is to assign the value pointed by a pointer to an integer, dereferencing should be used. If a function is meant to return an integer, it should not return a pointer. When dealing with arrays, remember that the array name behaves like a pointer to the first element, not an individual element of the array.

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[SOLVED] C - assigment makes integer from pointer without a cast warning

Thread: [solved] c - assigment makes integer from pointer without a cast warning, thread tools.

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I know it's a common error, and I've tried googling it, looked at a bunch of answers, but I still don't really get what to do in this situation.... Here's the relevant code: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char path[51]; const char* home = getenv( "HOME" ); strcpy( path, argv[1] ); path[1] = home; return 0; } -- there is more code in the blank lines, but the issue's not there (I'm fairly sure), so didn't see the point in writing out 100 odd lines of code. I've tried some stuff like trying to make a pointer to path[1], and make that = home, but haven't managed to make that work (although maybe that's just me doing it wrong as opposed to wrong idea?) Thanks in advance for any help

Re: C - assigment makes integer from pointer without a cast warning

path[1] is the second element of a char array, so it's a char. home is a char *, i.e. a pointer to a char. You get the warning because you try to assign a char* to a char. Note also the potential to overflow your buffer if the content of the argv[1] argument is very long. It's usually better to use strncpy.

Last edited by r-senior; March 10th, 2013 at 03:03 PM . Reason: argv[1] would overflow, not HOME. Corrected to avoid confusion.

Please create new threads for new questions. Please wrap code in code tags using the '#' button or enter it in your post like this: [code]...[/code].

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You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that.

Last edited by Christmas; March 10th, 2013 at 09:51 PM .

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Originally Posted by Christmas You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that. Excellent point. I've basically fixed my problem by reading up on pointers again (haven't done any C for a little while, so forgot some stuff), and doing: Code: path[1] = *home; the code doesn't moan at me when I compile it, and it runs okay (for paths which aren't close to 51 at least), but after reading what you read, I just wrote a quick program and found out that getenv("HOME") is 10 characters long, not 1 like I seem to have assumed, so I'll modify my code to fix that.

Yes, getenv will return the path to your home dir, for example /home/user, but path[1] = *home will still assign the first character of home to path[1] (which would be '/').

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What is Assignment Makes Pointer From Integer Without A Cast and How Can It Be Avoided?

Understanding the pointer-integer conundrum in programming.

In the realm of programming, particularly in languages like C and C++, pointers play a crucial role in managing memory and optimizing the performance of applications. However, they can also be a source of confusion and errors, especially for those new to these languages. One common error that programmers encounter is the “assignment makes pointer from integer without a cast” warning or error. This message can be perplexing, but understanding its meaning is essential for writing clean, efficient, and error-free code.

Decoding the Warning: Assignment Makes Pointer from Integer Without a Cast

The warning “assignment makes pointer from integer without a cast” is a compiler message that indicates a potential issue in your code. It occurs when an integer value is assigned to a pointer variable without explicitly casting the integer to a pointer type. This can happen for various reasons, such as mistakenly using an integer as a memory address or overlooking the need for a type conversion.

Why Does This Warning Matter?

Ignoring this warning can lead to undefined behavior in your program. Since pointers and integers are fundamentally different types, treating an integer as a memory address can cause your program to access memory locations it shouldn’t, leading to crashes, data corruption, or security vulnerabilities.

Examples of Pointer-Integer Assignment Issues

To better understand the warning, let’s look at some examples where this issue might arise:

In the example above, the variable num is an integer, and ptr is a pointer to an integer. The assignment of num to ptr without a cast triggers the warning.

Root Causes of the Pointer-Integer Assignment Warning

Several scenarios can lead to this warning. Understanding these scenarios can help you avoid the issue in your code.

• Accidental assignment of an integer to a pointer variable.
• Using an integer as a function return value where a pointer is expected.
• Incorrectly using an integer constant as a null pointer.
• Forgetting to allocate memory before assigning it to a pointer.

Strategies to Avoid Pointer-Integer Assignment Issues

To prevent the “assignment makes pointer from integer without a cast” warning, you can employ several strategies:

• Always use explicit casting when converting an integer to a pointer.
• Ensure that functions returning pointers do not accidentally return integers.
• Use the NULL macro or nullptr (in C++) for null pointers instead of integer constants.
• Properly allocate memory using functions like malloc or new before assigning it to pointers.

Using Explicit Casting

Explicit casting involves converting one data type to another by specifying the target type. In the context of pointers and integers, you can use a cast to tell the compiler that you intentionally want to treat an integer as a pointer.

Proper Function Return Types

When writing functions that are supposed to return pointers, ensure that the return type is correctly declared as a pointer type, and that the return statements within the function adhere to this type.

Using NULL or nullptr

Instead of using integer constants like 0 for null pointers, use the NULL macro in C or nullptr in C++ to avoid confusion and potential warnings.

Memory Allocation

Before assigning a pointer, ensure that it points to a valid memory location by using memory allocation functions.

Advanced Considerations and Best Practices

Beyond the basic strategies, there are advanced considerations and best practices that can help you write robust pointer-related code:

• Use static analysis tools to catch potential pointer-related issues before runtime.
• Adopt coding standards that discourage unsafe practices with pointers and integers.
• Understand the underlying architecture and how pointers are represented and used.
• Regularly review and refactor code to improve clarity and safety regarding pointer usage.

FAQ Section

What is a pointer in programming.

A pointer is a variable that stores the memory address of another variable. Pointers are used for various purposes, such as dynamic memory allocation, implementing data structures like linked lists, and optimizing program performance.

Why is casting necessary when assigning an integer to a pointer?

Casting is necessary because pointers and integers are different data types with different interpretations. A cast explicitly informs the compiler that the programmer intends to treat the integer as a pointer, which can prevent unintended behavior.

Can ignoring the pointer-integer assignment warning lead to security issues?

Yes, ignoring this warning can lead to security issues such as buffer overflows, which can be exploited by attackers to execute arbitrary code or cause a program to crash.

Is it always safe to cast an integer to a pointer?

No, it is not always safe. The integer should represent a valid memory address that the program is allowed to access. Arbitrary casting without ensuring this can lead to undefined behavior and program crashes.

What is the difference between NULL and nullptr?

NULL is a macro that represents a null pointer in C and is typically defined as 0 or ((void *)0). nullptr is a keyword introduced in C++11 that represents a null pointer and is type-safe, meaning it cannot be confused with integer types.

The “assignment makes pointer from integer without a cast” warning is a signal from the compiler that there’s a potential issue in your code. By understanding what causes this warning and how to avoid it, you can write safer and more reliable programs. Always be mindful of the types you’re working with, use explicit casting when necessary, and follow best practices for pointer usage. With these strategies in place, you’ll be well-equipped to handle pointers and integers in your coding endeavors.

For further reading and a deeper understanding of pointers, integer casting, and related topics, consider exploring the following resources:

• C Programming Language (2nd Edition) by Brian W. Kernighan and Dennis M. Ritchie
• Effective Modern C++ by Scott Meyers
• C++ Primer (5th Edition) by Stanley B. Lippman, Josée Lajoie, and Barbara E. Moo
• ISO/IEC 9899:201x – International Standard for Programming Language C
• ISO/IEC 14882:2017 – International Standard for Programming Language C++

These references provide a solid foundation for understanding the intricacies of pointers and type casting, as well as best practices for writing secure and efficient code in C and C++.

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Assignment makes pointer from integer without a cast

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Thread: Assignment makes pointer from integer without a cast

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I keep getting this error message: "warning: assignment makes pointer from integer without a cast" for line 17 (which I've made red). I've gotten this message a few times before, but I can't remember how I fixed it, and I can't figure out how to fix this one. Any suggestions? Also, it's a code that will take a password entered by the user and then run several for loops until it matches the password. It prints what it's figured out each time it guesses a new letter. Code: #include <stdio.h> #include <string.h> int main(void) { int i, j; char password[25]; char cracked[25]; char *p; char guess = '!'; printf("Enter a password of 25 characters or less: \n"); scanf("%s", password); printf("Password is being cracked..."); for (i = 0, p = password[i]; i < 25; i++, p++) { for(j = 0; j < 90; j++) { if (*p == guess) { strcpy(p, cracked); printf("\t %s \n"); break; } guess++; } //end <search> for loop } //end original for loop return 0; }

Last edited by deciel; 12-13-2011 at 01:57 AM .

Code: for (i = 0, p = password[i]; i < 25; i++, p++) password[i] is the value at index i of password . You want the address of said value, so you want p = &password[i] (or, equivalently, p = password + i ).

Oh! Thank you, it worked!

Originally Posted by deciel I keep getting this error message: "warning: for (i = 0, p = password[i]; i < 25; i++, p++) [/CODE] Note: password[i] == password[0] == *password since this is the assignment portion of for loop and i is set to zero (0).

If you want to set a pointer to the beginning of an array, just use Code: p = password An array name is essentially a pointer to the start of the array memory. Note, for a null terminated string, you could just test for Code: *p //or more explicitly *p == '\0' Also, a 25-element char array doesn't have room for a 25 character string AND a null terminator. And, ask yourself, what's going on when I enter, say a 10 character password, and i > 10.

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can't install cpan module Image::Magick

This question still has no solution (scroll down to "edit 5" to see the actual status).

I am trying to install Image::Magick on my headless Ubuntu server Ubuntu 14.04.2 LTS

Obviously the install-script is trying to include the header-file magick/MagickCore.h but can't find it.

When I search in search engines for magick/MagickCore.h then I find lots of postings where some people have the same problem and ask for, but instead of solution i did only find lots of answers saying "I have the same problem" - no solution.

Does anyone of you have a solution that works on a Ubuntu Server 14.04.2 ?

Edit 1: I already installed the package ImageMagic:

But I still get the error posted above.

Edit 2: Following the advise from one of the answers I also tried

followed by

And I still get the same error (script can't find magick/MagickCore.h)

Edit 3: I followed another advise and searched for magick/MagickCore.h . the answer was:

So I installed libmagickcore-dev (I exected sudo -i before):

This installation was successful, the previously missing header-file is now in the file system:

So I again tried

But I still get this error:

(» Datei oder Verzeichnis nicht gefunden « is German, my native language, and means » no such file or directory «)

Edit 4: Someone told me that the compiler is looking for magick/MagickCore.h not in /usr/include/ImageMagick/ but in /usr/include/ImageMagick-6/ . So I created a symbolic link:

and tried again to install the module:

So, the script did find magick/MagickCore.h , but now throws lots of other errors.

EDIT 5 (2015-09-17)

I've got the hint to install

apt-get install perlmagick

But it said, that the newest version already is installed. But I tried to install Image::Magick anyway. It still doesn't work:

This is the first error reported:

tmp-18091 for tmp-18091: Datei oder Verzeichnis nicht gefunden at /usr/share/perl/5.18/CPAN/Distribution.pm line 468.

which probably is in english:

tmp-18091 for tmp-18091: no such file or directory at /usr/share/perl/5.18/CPAN/Distribution.pm line 468.

• imagemagick

• I know this is old, and I don't know if it will help or even if you still have the issue, but ... I was having the same error, and found that the solution wasn't with cpan, but to install perlmagick: sudo apt-get install perlmagick After that, I could use Image::Magick in my Perl scripts. –  AntonH Sep 17, 2015 at 4:29
• 1 It is not old. I still can't install Image::Magick. I tried to install perlmagick, but i got the message, that I still have the newest version installed. –  Hubert Schölnast Sep 17, 2015 at 10:24
• 1 I'm sorry I can't help more :( I had checked my installation of ImageMagick with sudo apt-get install imagemagick and it told me I had latest version installed (version with Ubuntu 14.04 desktop; I don't know if version between server and desktop is different, or if server doesn't provide dev files). Didn't install and PHP5 modules, and I couldn't find the ImageMagick.h files on my system. Good luck. If I find something I'll check back in. –  AntonH Sep 17, 2015 at 17:15

5 Answers 5

Diego's solution works.

A similar procedure has worked for me for versions ImageMagick-6.9.1 and ImageMagick-6.9.5.

Build Image::Magick perl module from source

Go to: http://www.imagemagick.org/script/download.php

Download a tar.gz file from one of the mirrors:

Extract into ~/home/src:

Configure Image Magic with Perl Bindings

Make sure perl dev files are installed

Install Image Magick

Check that it works:

My final solution installing Image magick was:

(After trying all your steps)

Download current version of ImageMagick from the source

6.9.2 in my case:

After the download, switch to the Download folder and:

and then all goes smoothly

(*) another missing part, lperl , so in the process I have to add: sudo apt-get install libperl-dev

Install the package php5-imagick and try installing via CPAN again.

• I'm not using php on my server, but I followed you advise and installed php5-imagick. The result: nothing changed. The installation of the cpan-modul still throws the same error. –  Hubert Schölnast May 22, 2015 at 19:54
• Thanks for trying. I've had CPAN issues that relied on my installing PHP packages before. Sorry it didn't help in this case. My only other recommendation would be to upgrade CPAN itself via CPAN. –  earthmeLon May 22, 2015 at 19:56
• I executed cpan-outdated -p | cpanm immediately before I tried to install Image::Magick. I think this should also have updated cpan itself. –  Hubert Schölnast May 22, 2015 at 20:02
• It's looking for imagemagick devel headers - did you verify you have those installed? –  Thomas Ward ♦ May 23, 2015 at 21:47

Try to install headers:

Along with -dev packages:

Obviously the perl-lib is less thoroughly maintained than the Ubuntu repos.

Fixed it for while cpanm failed.

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